Denote $ S(x,y,z) = \sum{x^2(y+z)} = \sum{x^2 - x^3}$. Note that $ S$ is symmetric. Since $ \frac{d^2}{dx^2}(x^2 - x^3) = 2 - 6x$, $ x^2 - x^3$ is convex in $ [0,\frac 13]$ and concave in $ [\frac 13,1]$. For three number $ x,y,z$, two of them must be both greater than $ \frac 13$ or both less than or equal to $ \frac 13$. Because of symmetry, we assume that they are $ y,z$. Hence, we must have: $ S(x,y,z) \leq S(x,y+z,0)$ or $ S(x,y,z) \leq S(x,\frac{y+z}{2},\frac{y+z}{2})$ Combining both, we have \[ S(x,y,z) \leq \max(S(x,y+z,0), S(x,\frac{y+z}{2},\frac{y+z}{2}))\] Arithmetic Compensation Method gives us: $ max S = \max (S(1,0,0) , S(\frac 12, \frac 12, 0), S(\frac 13,\frac 13,\frac 13)) = \max(0,\frac 14, \frac 59) = \frac 59$. The maximum is attained at $ x = y = z =\frac 13$.

April wrote: If $ x$, $ y$, $ z$ are nonnegative real numbers with the sum $ 1$, find the maximum value of $ S = x^2(y + z) + y^2(z + x) + z^2(x + y)$ and $ C = x^2y + y^2z + z^2x$. \[ a^2 b + b^2 c + c^2 a + abc \le \frac{4}{{27}}\left( {a + b + c} \right)^3 \]

also http://www.mathlinks.ro/Forum/post-768.html#768

$ \sum x^2(y + z)\leq\frac {1}{4}\Leftrightarrow\sum x^2(y + z)\leq\frac {1}{4}(\sum x)^3$ $ \Leftrightarrow(\sum x)^3\geq 4\sum x^2(y + z)\Leftrightarrow\sum x^3 + 6xyz\geq\sum x^2(y + z)$ It's true by Schur's ineq

Honey_S wrote: April wrote: If $ x$, $ y$, $ z$ are nonnegative real numbers with the sum $ 1$, \[ a^2 b + b^2 c + c^2 a + abc \le \frac {4}{{27}}\left( {a + b + c} \right)^3 \] Find the greatest value of $ a^nb+b^nc+c^na$ where $ n\in N$

the generalization has also been discussed http://www.mathlinks.ro/Forum/viewtopic.php?t=2549